Let $X$ be a finite set. Prove that every ultrafilter is a point filter. 
Let $X$ be a finite set. Prove that every ultrafilter is a point
  filter.

My attempt:
Let $\mathcal{U}$ be an ultra filter. Write $X = \bigcup_{i=1}^n \{x_i\} \in \mathcal{U}$. Then there exists $i \in \{1,2, \dots, n\}$ such that $\{x_i\} \in \mathcal{U}$, because it is an ultrafilter. I now claim that $\mathcal{U}$ is the point filter of the point $x_i$, i.e. we have to prove that $\mathcal{U} \subseteq \{F \subseteq X \mid x_i \in F\}$ and equality will follow because it is an ultrafilter.
How to proceed?
 A: Let $\mathcal{F}$ be a filter on a set $X$ such that there is some $x \in X$ with $\{x\} \in \mathcal{F}$. Then
$$\mathcal{F} = \mathcal{F}_x := \{A \subseteq X: x \in A\}$$.
The proof is immediate: if $F \in \mathcal{F}$ then as $\{x\} \in \mathcal{F}$, $\{x\} \cap F \neq \emptyset$ as any two sets in a filter intersect in a member of the filter and all filter members are non-empty (two axioms of filters!). But that just says that $x \in F$ and thus $F \in \mathcal{F}_x$.
On the other hand, if $F \in \mathcal{F}_x$, then $x \in F$ or $\{x\} \subseteq F$ and as $\{x\} \in \mathcal{F}$, $F \in \mathcal{F}$, as filters are closed under supersets (another filter axiom!).
So we have equality of these filters.
A: You've already shown that $\mathcal{U}$ contains a singleton $\{x_i\}$. Now observe that the filter generated by $\{x_i\}$ is an ultrafilter and must be contained in $\mathcal{U}$. But ultrafilters are maximal filters with respect to $\subseteq$, so in fact $\mathcal{U}$ is the filter generated by $\{x_i\}$.
